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Survey accuracy A study investigating the relationship between age and annual medical expenses randomly samples 100 individuals in Davis, California. It is hoped that the sample will have a similar mean age as the entire population. a. If the standard deviation of the ages of all individuals in Davis is \(\sigma=15,\) find the probability that the mean age of the individuals sampled is within two years of the mean age for all individuals in Davis. (Hint: Find the sampling distribution of the sample mean age and use the central limit theorem. You don't have to know the population mean to answer this, but if it makes it easier, use a value such as \(\mu=30 .\) ) b. Would the probability be larger, or smaller, if \(\sigma=10 ?\) Why?

Short Answer

Expert verified
a. Probability is 0.8164. b. Probability is larger with \(\sigma=10\) because of a smaller SE.

Step by step solution

01

Understand the Central Limit Theorem

The Central Limit Theorem (CLT) states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough (usually n > 30). In this exercise, it helps us determine the distribution of the sample mean.
02

Define the Sampling Distribution

Since the sample size is n = 100, the sampling distribution of the mean age of the sample will be normal with mean \( \mu \) and standard deviation \( \sigma / \sqrt{n} \). Given \( \sigma = 15 \), the standard deviation of the sample mean, known as the standard error (SE), is \( SE = 15 / \sqrt{100} = 1.5 \).
03

Calculate Z-Scores for the Range Within Two Years

We are asked for the probability that the sample mean is within two years of the population mean. This means we are interested in the range \( \mu - 2 \leq \bar{x} \leq \mu + 2 \). The Z-scores corresponding to these limits are calculated as: \[ Z = \frac{(\bar{x} - \mu)}{SE} = \frac{2}{1.5} \approx 1.33 \]So the Z-scores are \(-1.33\) and \(1.33\).
04

Use the Standard Normal Distribution

Using the Z-table, find the probabilities for the Z-scores. The probability of a Z-score being less than 1.33 is approximately 0.9082, and the probability of a Z-score being less than -1.33 is 0.0918. The probability that the sample mean falls within this range is:\[ P(-1.33 < Z < 1.33) = P(Z < 1.33) - P(Z < -1.33) = 0.9082 - 0.0918 = 0.8164 \]
05

Analyze the Effect of Smaller Standard Deviation (\(\sigma=10\))

If \( \sigma \) decreases to 10, the standard error becomes \( SE = 10 / \sqrt{100} = 1 \). Repeating Steps 3 and 4 with SE = 1:\[ Z = \frac{2}{1} = 2 \]So the Z-scores are \(-2\) and \(2\). The probability of Z between -2 and 2 is:\[ P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544 \]The probability is larger because the variance in individual ages decreases, leading to a smaller standard error, which makes the sample mean more likely to be close to the population mean.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sampling Distribution
When we deal with samples, we're usually interested in understanding how well a sample mean, like the average age from our survey, represents the true population mean. This is where the concept of sampling distribution comes into play. The sampling distribution of a sample mean is essentially the distribution of all possible sample means that we could obtain by taking repeated samples from the population.

It provides insight into how these sample means would behave if we could, in theory, take an infinite number of samples. This helps us predict the variability and distribution of sample means. A key takeaway here is that by using the Central Limit Theorem, we can expect the sampling distribution to be approximately normal if the sample size is sufficiently large (typically n > 30).
  • This approximation to normality makes it easier to perform hypothesis testing.
  • Helps in estimating probabilities related to sample means, as we did with ages in Davis.
Understanding this distribution is crucial for making predictions and inferences about the population from which our sample is drawn.
Standard Deviation
The standard deviation, denoted as \( \sigma \), is a measure of how much individual data points typically deviate from the mean of a set of numbers. In our context, it's used to quantify the variability in ages among all individuals in Davis.

A larger standard deviation indicates that data points are spread out over a broader range of values. Conversely, a smaller standard deviation means data points tend to be closer to the mean. For our study, knowing this value allows us to calculate the standard error, which in turn impacts the precision and reliability of our sample mean.
  • The natural variability in age can be captured by the standard deviation.
  • Affects our confidence in estimating the true mean age of the larger population.
Remember that the standard deviation is a foundational concept in understanding variability and uncertainty in data analysis.
Z-scores
Z-scores are a way of standardizing the values in a data set to understand how far and in what direction they deviate from the mean, measured in standard deviations. In the case of our exercise, we used Z-scores to determine how far the sample mean age was from the population mean age.

By converting our bounds (within two years of the mean age) into Z-scores, we were able to determine the probability of the sample mean falling within that range. This is done by calculating the Z-score for a given value, which is performed using the formula:
\[ Z = \frac{{(\bar{x} - \mu)}}{SE} \]
  • Z-scores help us locate our data point on the standard normal distribution.
  • They are essential for finding probabilities and making inferences about sample data.
Z-scores are versatile tools for understanding the relative position of a sample mean in the context of a larger, normal distribution.
Standard Error
The standard error is a crucial concept when analyzing sample data since it's a way to measure how much the sample mean would differ, on average, from the true population mean. Essentially, it's the standard deviation of the sampling distribution of the sample mean.

In our example, with a sample size of 100 and a population standard deviation \( \sigma = 15 \), we calculate the standard error \( (SE) \) as follows: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{100}} = 1.5 \]
  • The smaller the standard error, the more accurate our estimate of the population mean.
  • A lower standard error implies that our sample mean is a precise estimate of the true mean.
This concept is pivotal when it comes to determining how much we can trust the mean of our single sample to reflect the overall population mean. It decreases with larger sample sizes, increasing our confidence in the sample mean's representativeness.

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Most popular questions from this chapter

Relative frequency of heads Construct the sampling distribution of the sample proportion of heads obtained in the experiment of flipping a balanced coin: a. Once. (Hint: The possible sample proportion values are \(0,\) if the flip is a tail, and \(1,\) if the flip is a head. What are their probabilities?) b. Twice. (Hint: The possible samples are \(\\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH},\) \(\mathrm{TT}\\}\), so the possible sample proportion values are 0 , \(0.50,\) and \(1.0,\) corresponding to \(0,1,\) or 2 heads. What are their probabilities?) c. Three times. (Hint: There are 8 possible samples.) d. Refer to parts a-c. Sketch the sampling distributions and describe how the shape is changing as the number of flips \(\mathrm{n}\) increases.

Other scenario for exit poll Refer to Examples 1 and 2 about the exit poll, for which the sample size was \(3889 . \mathrm{In}\) that election, \(40.9 \%\) voted for Whitman. a. Define a binary random variable \(X\) taking values 0 and 1 that represents the vote for a particular voter \((1=\) vote for Whitman and \(0=\) another candidate \()\) State its probability distribution, which is the same as the population distribution for \(X\). b. Find the mean and standard deviation of the sampling distribution of the proportion of the 3889 people in the sample who voted for Whitman.

Multiple choice: CLT The central limit theorem implies a. All variables have approximately bell-shaped data distributions if a random sample contains at least about 30 observations. b. Population distributions are normal whenever the population size is large. c. For sufficiently large random samples, the sampling distribution of \(\bar{x}\) is approximately normal, regardless of the shape of the population distribution. d. The sampling distribution of the sample mean looks more like the population distribution as the sample size increases.

True or false \(\quad\) As the sample size increases, the standard deviation of the sampling distribution of \(\bar{x}\) increases. Explain your answer.

Blood pressure Vincenzo Baranello was diagnosed with high blood pressure. He was able to keep his blood pressure in control for several months by taking blood pressure medicine (amlodipine besylate). Baranello's blood pressure is monitored by taking three readings a day, in early morning, at mid-day, and in the evening. a. During this period, the probability distribution of his systolic blood pressure reading had a mean of 130 and a standard deviation of \(6 .\) If the successive observations behave like a random sample from this distribution, find the mean and standard deviation of the sampling distribution of the sample mean for the three observations each day. b. Suppose that the probability distribution of his blood pressure reading is normal. What is the shape of the sampling distribution? Why? c. Refer to part b. Find the probability that the sample mean exceeds \(140,\) which is considered problematically high. (Hint: Use the sampling distribution, not the probability distribution for each observation.)

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